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I am having problem to see the contradiction of the following working:

Consider two arbitrary operators $\hat A$, $\hat B$, such that $[\hat A, \hat B ] = c\hat1$, where $c$ is a non-zero constant. If we further assume $\hat A$ is hermitian and obey the eigenvalue equation given by $$A\left|a\right> = \lambda_a\left|a\right>,$$ where $\lambda_a$ is real number, consider the expression $\left<a\right|[\hat A, \hat B ]\left|a\right>$:

On the one hand, $\left<a\right|[\hat A, \hat B ]\left|a\right> = \left<a\right|c\hat1\left|a\right> = c\left<a|a\right> \neq 0$,

but if we expand the commutator, $\left<a\right|[\hat A, \hat B ]\left|a\right> = \left<a\right|\hat A \hat B - \hat B \hat A\left|a\right> = (\left<a\right|\hat A) \hat B \left|a\right> - \left<a\right|\hat B (\hat A\left|a\right>) = \lambda_a\left<a|B|a\right> - \lambda_a\left<a|B|a\right> = 0$.

I could not see why these two results are different, I assume there's some properties of the associativity with respect to the commutator I am missing, but I can not figure it out formally.

mike stone
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Zeqi Zhang
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    I believe this won't change the main point, but shouldn't the $\langle a \vert a \rangle$ actually read $\langle a \vert \hat{B} \vert a \rangle$ on the final steps of the calculation? – Níckolas Alves Dec 19 '21 at 22:01
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    I am not sure, and hence I'm just commenting instead of posting an answer, but I believe the problem you're having has to do with the fact that these operators cannot be bounded, and hence you might be running on some issues with their domains of definition (an example would be $\hat{p}$ and $\hat{x}$, but their "eigenstates" are not really states, since they are not normalizable). See, e.g., this entry on Wikipedia – Níckolas Alves Dec 19 '21 at 22:20
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    Possible duplicate: https://physics.stackexchange.com/q/418587/2451 – Qmechanic Dec 19 '21 at 22:29

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